Requiem for Relativity

Shift to a pole with yearly torque-free precession is energetically possible for Earth

(This post has been extensively rewritten two months after originally posted. I corrected a significant typographical error in my original computer program. Also I have, perhaps, slightly increased the accuracy of my analysis, by considering Earth's core/mantle structure; I had discussed my analysis with Dr. Ken Driessel, an expert in classical mechanics, who warned that the core/mantle structure might be significant.)

Using data from the 1975 Columbia Encyclopedia (Fig. A, article "Earth") I sometimes will approximate Earth as having a liquid core (the small solid core inside this is without effect) of density 11.0 gm/cm^3 and radius 3500km, and a mantle of density 4.425 gm/cm^3, so the entire Earth has radius 6400km and average density 5.5 gm/cm^3. When Earth undergoes torque-free precession, with its spheroid axis, say, 30deg different from its rotation axis, but the amount of spheroidal flattening about the same, then surface points will find themselves as much as 1/2 radian nearer or farther from the pole than is proper for the amount of bulge at that point. At temperate latitudes this amounts to +/- 6000/300/2 = 10 km of bulge. If this were compensated only by ocean movement, then at most 10 / 2.5 = 4 km of increased ocean depth suffices to equalize surface potential, because the density of water (=1) is almost negligible compared to the average density of Earth (=5.5) and according to the textbook formula (e.g., A. S. Ramsey, "Newtonian Attraction", sec. 7-42) elevation away from an unchanging spheroid is 2.5x as effective, at changing a point's potential, as is elevation caused by additional bulging of the spheroid under the point. Generally, stratigraphic evidence of thousands of yearly 4 km = 13,000 ft rises in ocean level, at temperate latitudes, is lacking. So, the rise might mainly have been provided by movement of crustal rock (density ~2.2) or hydrocarbon layers (density < 1). Thus my theory differs from Hapgood's: he required ~3000km horizontal motions of the crust, to cause pole shift; I require only smooth ~10km vertical motions of it, to reduce flooding when torque-free precessional pole shift occurs (atmospheric pressure would change little, because the atmosphere also seeks the equipotential surface).

Earth's dense core causes Earth's "actual oblateness" k0=1/297, to differ from Earth's "equivalent oblateness" k=1/304, i.e. the oblateness of a homogeneous spheroid having Earth's mass, radius and moments of inertia:

k = ( 6400^5 * 4.425*k0 + 3500^5 * (11.0-4.425)*1/462 ) /
(6400^5 * 4.425 + 3500^5 * (11.0-4.425))

and the converse formula giving k0 from k, is also linear. The core's bulge dr/r = 1/462 is found by equating omega^2 * r^2 / 2 to G*11.0*4/3*pi * r^3/r^2 * dr. (Likewise Earth's bulge would be 1/231 if Earth were homogeneous with density 11.0/2 = 5.5; the actual bulge 1/297 is less, because as discussed above, movement away from a dense core, is more efficient at equalizing potential, than is distortion of a homogeneous spheroid.)

From Earth's "equivalent oblateness" and the online NASA Fact Sheet figure, 8.034*10^44 in cgs units, for Earth's polar-axis moment of inertia, I find the moment of inertia for a sphericized Earth. Then by subtracting the moment of inertia of the core, I find the moment of inertia, i0, for the sphericized mantle. Without correcting oblateness, only subtracting the core's part, I find Earth's mantle's kinetic energy of rotation.

Conservation of energy and of angular momentum, give two equations relating the spheroidal mantle's large & small moments of inertia, i3 & i1, resp., and its spin components parallel & perpendicular to the spheroidal axis, wz & A*wz, resp. (see Goldstein, "Classical Mechanics", sec. 5-6). Because the oblateness is small, constant volume implies approximately i3 + 2*i1 = 3*i0 (I used an exact formula for i1 as a function of i0 & i3, for a homogeneous spheroid, but the exact result hardly differs from that found from the simple formula). The fourth needed equation is (i3 - i1) / i1 = (frequency of tropical year) / wz (see Goldstein).

The conservation of energy equation which I adopt from Goldstein, requires knowledge of the energy liberated as Earth assumes a more spherical shape; this must be added to the kinetic term. I use Earth's gravitational self-energy 2.4884*10^39 erg (Marchenko & Zayats, Stud. Geophys. Geod. 55:35-54, 2011, "Conclusions", p. 52, based on the "piecewise continuous PREM density model"). If Earth were a homogeneous spheroid, the self-energy due to oblateness would equal the total self energy, times 4/5*(k/3)^2, seen by expanding

1 / sqrt((x*(1+k0/3))^2+(y*(1+k0/3))^2+(z*(1-2*k0/3))^2)

in k0, and averaging on the sphere x^2 + y^2 +z^2 = 1. Estimating that 17.9% of Earth's self-energy is core-core self energy, and assuming that the oblateness of the core does not change, I proceeded by lessening the amount given by the above formula, by 17.9%.

The self-energy, above, released by decreasing oblateness, becomes doubled because Earth isn't really incompressible, but instead resembles an adiabatic monatomic gas P * V^gamma = const, where gamma=5/3 and V is proportional to r^3; P = P0/r^4, because both the gravitational force and the reciprocal of the area to which it is applied, are proportional to r^2. Thus P0 * r = const. When decreased oblateness decreases interparticle distances by, say, 1%, then forces and pressure and P0 increase 2%, so r must decrease by 2%, giving in our situation another 2*4/5*(k/3)^2 term of liberated energy. According to the virial theorem, half this becomes increased thermal motion of the particles, and half is available mechanically, giving a grand total (1 + 2/2) * 4/5*(k/3)^2 (times (1 - 0.179), to account for the univolvement of Earth's core in the torque-free precession).

By time-averaging the centrifugal potential at pole and equator, or else by time-averaging the shapes of the equilibrium spheroids, I find that for torque-free precession at an angle alpha, Earth's oblateness becomes less by a factor 1 - 1.5*sin^2(alpha). If I multiply the liberated self-energy by a factor of 1.022, i.e. only 2% different from my estimate, I find that:

1. Torque-free precession with period one year, occurs for alpha = 18.90deg; i.e. the pole moves along 71deg latitude. Hapgood's determinations of the last three Ice Age poles, average latitude 65.0 +/- SEM 3.6, range 60-72deg. So the agreement with Hapgood's poles is acceptable.

2. Earth's actual oblateness agrees with the averaging formula 1 - 1.5*sin^2(alpha), discussed above.

In other words, for torque-free precession to occur at latitudes other than Hapgood's determination for Ice Age poles, Earth's oblateness must assume an implausible value. Only when torque-free precession occurs along the 71st parallel, is Earth's oblateness what it should be according to the law of centrifugal force. Furthermore the energy required for this, is just what I estimate from the two equal terms, due to Earth's loss of oblateness and to Earth's resulting slight shrinkage as a monatomic adiabatic gas.

A good approximation, is simply to solve 1 - 1.5*sin^2(alpha) = 304/365, for alpha, where Earth's equivalent oblateness, 1/304, is used for better accuracy than Earth's actual oblateness, 1/297. The answer, 19.50deg, agrees well with the result of the more detailed calculation, involving Earth's core, above.

A wandering pole at a large precession angle, would give the sun a more dramatic motion: resembling our temperate summers but our arctic nights. This makes some ancient statements about solar motion seem less exaggerated.

Dr Joe, I don't pretend to understand your thesis, but I try and now I wonder what you mean by gravity energy of several kinds. Gravity is a force in my world having nothing in common with energy. Can you explain how you get energy from gravity?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />...energy from gravity?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

This is my cue to restate my hypothesis in everyday language. When a spinning skater pulls in his arms, that requires energy (the chemical energy he gets from his food). The energy from pulling in the arms, becomes the increased kinetic energy of rotation.

Likewise, if Earth - either with no change in shape, or with a change toward being more perfectly spherical - rotates about an axis that is someplace away from the pole of the oblate spheroid, then the mass, of the Earth, is, on the average, slightly closer to the axis of rotation. Like the skater, the Earth must rotate slightly faster if the angular momentum is to remain constant. It turns out that the kinetic energy of such rotation is slightly greater than before. Where does this kinetic energy come from? It can come from the "falling" of the Earth into a slightly rounder shape.

If one fixed axis were replaced by another fixed axis, Earth eventually would become an oblate spheroid congruent to the old oblate spheroid, but with a different pole, a pole at the new rotation axis. But under my hypothesis, the axis precesses. So, the flattening effect of each axis position is averaged, and the average of all these oblate spheroids, is a spheroid that is less oblate and with the same pole (geometric pole, not rotation pole) as before. As Earth falls into a more spherical shape, the gravitational potential energy released, becomes the increased kinetic energy of rotation that is needed to conserve angular momentum and make the new state of affairs possible.

It turns out that there is enough gravitational potential energy released, to establish the new state as a precession of Earth's axis along the 69th parallel. This is in good agreement with the position of the three last Ice Age poles according to the geological estimates collected by Hapgood; and in good agreement with the precession rate needed to synchronize with Earth's tropical year, thereby having a climatic effect similar to an outright new fixed axis. (Due to conservation of momentum, the axis is always about the same direction in space - except for astronomical precession - but Earth shifts under the axis so that the position of the axis, relative to Earth, changes grossly, under my hypothesis.)

Earth's area exceeds the area of a sphere of equal volume, by 2.02 parts per million; see oblate spheroid area formula, in CRC Math Tables mensuration formulas, b=a*296/297, e=sqr(1-(b/a)^2). If Earth were less oblate - the oblateness I predict for the torque-free axis precession mode with 1/365 flattening - the excess area would be only 1.34 ppm. So Earth would find itself with an excess skin of 0.68 ppm. This corresponds to a dike 68 cm wide along one edge of a square 1000km on a side, for example. It would entail a significant, but believable and not unprecedented amount of seismic activity.

IOW, this is not enough to account for a Dec 21, 2012 event? But it might be enough to rid us of California.

I realize that you have only calculated area changes above, but if this dike turned out to be 1000 meters tall instead of 68 cm tall (dikes are seldom taller than they are wide, and you did not specify a height so I just made a guess), the implications would likely be a little different. Any idea which it is more likely to be?

Dr Joe, Thank you for this detailed description about how the surface of Earth excess area due to angular momentum. I don't see how this gets energy enough to cause ice to form and remain for 90,000 years or so. I assume that is way you did this--

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Larry Burford</i>
<br />...if this dike turned out to be 1000 meters tall...
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Consider many concentric spheroids all with the same flattening, which all then become spheres without changing their internal volumes. All of them will have the same proportion of excess area, i.e. wrinkled skin, but these wrinkles needn't be in the same place. The height of the "dike", i.e. the height of the wrinkle in the uppermost spheroid, depends on the mechanical properties of the material and the strength of the connection between the spheroids. It needn't be very high. My method of calculation only tells me the area of the skin overlap, not its depth.